TMM4175 Polymer Composites

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CASE STUDY: Transformation by substitution of axes¶

Problem statement: Strains, stresses and engineering constants are given in a coordinate system 1,2,3 illustrated in Figure-1. Transform strains, stresses and enginnering constants to the coordinate system x,y,z.

Figure-1

Strains¶

The transformation of normal strains are straight forward by simple substituion of axes and therefore, indices:

\begin{equation} \varepsilon_{x}=\varepsilon_{3}, \quad \varepsilon_{y}=\varepsilon_{1}, \quad \varepsilon_{z}=\varepsilon_{2} \end{equation}

For the shear strains we must pay slightly more attention. Figure-2 (b) shows a deformed body representing, by definition, a positive shear strain $\gamma_{13}$. The same body from the same perspective is illustrated in Figure-2 (a) with the axes y and x. Now, flipping the axes and the deformed body as illustrated in Figure-2 (c), and we can draw the conclusion:

\begin{equation} \gamma_{yx}=-\gamma_{13} \end{equation}

Figure-2

The same logic applies for the relation between $\gamma_{yz}$ and $\gamma_{12}$: \begin{equation} \gamma_{yz}=-\gamma_{12} \end{equation}

while the remaining shear strain is directly related:

\begin{equation} \gamma_{zx}=\gamma_{23} \end{equation}

Since $\gamma_{ji}=\gamma_{ij}$, the summary can be written as

\begin{equation} \begin{bmatrix} \varepsilon_{x} \\ \varepsilon_{x} \\ \varepsilon_{x} \\ \gamma_{yz} \\ \gamma_{xz} \\ \gamma_{xy} \end{bmatrix}= \begin{bmatrix} \varepsilon_{3} \\ \varepsilon_{1} \\ \varepsilon_{2} \\ -\gamma_{12} \\ \gamma_{23}\\ -\gamma_{13} \end{bmatrix} \end{equation}

Stresses¶

The arguments made for strains can easilly be adapted to stresses as well. Hence,

\begin{equation} \begin{bmatrix} \sigma_{x} \\ \sigma_{x} \\ \sigma_{x} \\ \tau_{yz} \\ \tau_{xz}\\ \tau_{xy} \end{bmatrix}= \begin{bmatrix} \sigma_{3} \\ \sigma_{1} \\ \sigma_{2} \\ -\tau_{12} \\ \tau_{23}\\ -\tau_{13} \end{bmatrix} \end{equation}

Engineering constants¶

\begin{equation} E_x=E_3 \\ E_y=E_1 \\ E_z=E_2 \\ G_{xy}=G_{31}=G_{13} \\ G_{xz}=G_{32}=G_{23} \\ G_{yz}=G_{12} \\ \nu_{xy}=\nu_{31}=\nu_{13} \frac{ E_3 }{E_1} \\ \nu_{xz}=\nu_{32}=\nu_{23} \frac{ E_3 }{E_2} \\ \nu_{yz}=\nu_{12} \end{equation}

Remember that

\begin{equation} G_{ji}=G_{ij} \end{equation}

while

\begin{equation} \frac{ \nu_{ji} }{E_j} = \frac{ \nu_{ij} }{E_i} \end{equation}

Left as exercise: Verify the specific transformation in this case study using the relations for general transformation.

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